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Urgent help with calculus question pls?

Q1) https://gyazo.com/52b8da874d0d710cff49aa5d8bfdc8fa

a) I don't know what to do here. I know da/dx = 0 at a stationery point but
since they don't want you to do d^2a/dx^2 because it would be ugly, I'm left
with taking first principles so when I subbed in when x = sqroot(50) and got
da/dx = 0. I know if I take a value of x either on left or right of it and get a positive, 0,
negative value then I'll have a maximum but firstly 1) I don't know what exact value to
take since the graph changes really quickly as shown in the right picture.

b) I have no clue on how to get the values x, which gives the minimum area.
I tried splitting the quadrilateral into 2 right angle triangles. I know that
(8.6)/2 = 24 .*. the other triangle of [x.sqrt(100-x^2)]/2 = 24 but don't know if that method
I just mentioned just now is the right method of getting the minimum values of x and if so
Idk how to get the values of x by solving it here '[x.sqrt(100-x^2)]/2 = 24'.

c) I don't understand this. I thought x = can't be a negative value as a length can't be negative
or can area be negative so its not possible or that the area doesn't exist/area at this minimum
is not possible or something? Any help would be appreciated.

Q2) https://gyazo.com/e8224f6d3ab1eefa24d56d4b9b4bba6a

Q3) https://gyazo.com/7185f478b8911163045ee7daec5e8e60

Q4) https://gyazo.com/b45290bd5a3c0785dc59165466359a77

I would appreciate it if someone could guide me on how to solve the ones above as I tried asking lecturers/workshops and class mates but nobody knows and are stuck themselves. I just don't want to be stuck and would appreciate if someone could help me. :smile:

(edited 7 years ago)

Reply 1

DFranklin

Q1 is honestly so badly posed that it's no wonder no-one is sure what to do. (Part (a) is straightforward, just use the plot as it tells you to. But the "meaning" of x < 0 doesn't seem tractable, and it's unclear what the distinction between (b) and (c) is supposed to be either).

Reply 2

atsruser

Original post by XxKingSniprxX

They told you to look at the graph to check that it's a maximum...

b) I have no clue on how to get the values x, which gives the minimum area.

Just look at the graph. The next question answers this one anyway, so I don't see why they have asked it.

c) I don't understand this. I thought x = can't be a negative value as a length can't be negative
or can area be negative so its not possible or that the area doesn't exist/area at this minimum
is not possible or something? Any help would be appreciated.

This is a very interesting question. The only answer that makes sense to me is this:

a) We need to assign a sign to x - we can do this by saying that lines drawn clockwise from the bottom left vertex are +ve. i.e. we draw arrows on each edge of the quadrilateral so that they indicate an anticlockwise turn. Draw a anticlockwise circular arrow inside the quad. to indicate this.

b) Now we can reduce x to 0 by moving the bottom right vertex closer to the bottom left clockwise along the circumference - when they meet we have a triangle of sides +6, +8, +10 which is right angled with area 24, which corresponds to the value of A(0) on the curve.

c) We continue to move the bottom right vertex clockwise along the circumference so that x is non-zero again. We now have formed two triangles. Assign a direction to x by noting which way its arrow must go to conform with the arrows on the 6 and

\sqrt{100-x^2}

lines - you will see that it gets a clockwise arrow which indicates a -ve length.

d) Draw a circular arrow inside the newly formed bottom triangle - you will see that it goes clockwise - this indicates that it can be considered to have -ve area. The other triangle still have +ve area so when we add the areas of the triangles, we are reducing the total (signed) area of the shape.

e) At some point as we make x more and more -ve, we get two triangles one with area +A, one with area -A to give a total of 0.

This probably will make more sense if you have come across the idea of calculating signed area of shapes via determinants. It would likely make more sense if I could draw pictures here too.